Molecular Shapes
Presentation
•
Science
•
10th Grade
•
Practice Problem
•
Hard
+7
Standards-aligned
Denise Fanelli
FREE Resource
23 Slides • 6 Questions
1
Reorder
Place the steps for drawing Lewis structures in the correct order
(FYI you can scroll down on the answer choices to read the whole thing)
Add the total number of valence electrons
Determine the central atom
Draw remaining atoms around the central atom and draw bonds
Fill in lone pairs on the outer atoms, then place any remaining electrons on the central atom.
Check your work, add double/triple bonds if needed.
2
Hotspot
Click on the atom that makes this Lewis structure incorrect
3
Hotspot
Click on the atom that makes this Lewis structure incorrect.
4
Trigonal Planar – AX3
�
The formula is AX3 and an example would be SO3
�
It tells us there are three bonds to the central atom
�
We start with the Lewis structure for SO3
There are 4 BP and 0 LP of e-
around the central atom!
�
These central atom e- tell us SO3has a trigonal planar geometry
with bond angles of 120⁰
It is 120⁰
(1/3 of a circle)
from bond to bond
S
O
O
O
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23
Multiple Choice
What type of molecular geometry and bond angles would SO₃ have?
(You'll need to draw the Lewis structure, determine the AXE, then compare of the chart on the back of your periodic table.)
Linear, 180°
Trigonal planar 120°,
Trigonal pyramidal, 107°
Tetrahedral, 109.5°
24
25
Multiple Choice
What type of molecular geometry and bond angles would OCl₂ have?
(You'll need to draw the Lewis structure, determine the AXE, then compare of the chart on the back of your periodic table.)
Linear, 180°
Bent, 105°
Trigonal pyramidal, 107°
Tetrahedral 109.5°
26
27
Multiple Choice
What type of molecular geometry and bond angles would CBr₄ have?
(You'll need to draw the Lewis structure, determine the AXE, then compare of the chart on the back of your periodic table.)
Trigonal pyramidal, 107°
Tetrahedral, 109.5°
Bent, 105°
Linear, 180°
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29
Practice Problem
What geometry and bond angle would CBr4 have?
CBr4
There are 32 total valence e-:
C: (1 atom)(4 valence e-) = 4 e-
Br: (4 atoms)(7 valence e-) = 28 e-
C
There is only one carbon atom, so it is our central atom!
Fill in bonding e- first!
Fill the outermost atoms next
All of the e- have
been used and all of
the atoms are
happy with 8 e-
Br
Notice there are 4 BP and 0 LP of e- around the central atom!
Br
4 BP & 0 LP tells us it is:
tetrahedral w/ bond
angles of 109.5⁰
Br
Br
Place the steps for drawing Lewis structures in the correct order
(FYI you can scroll down on the answer choices to read the whole thing)
Add the total number of valence electrons
Determine the central atom
Draw remaining atoms around the central atom and draw bonds
Fill in lone pairs on the outer atoms, then place any remaining electrons on the central atom.
Check your work, add double/triple bonds if needed.
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